3.1.34 \(\int \frac {e^{4 i \text {ArcTan}(a x)}}{x^4} \, dx\) [34]

Optimal. Leaf size=62 \[ -\frac {1}{3 x^3}-\frac {2 i a}{x^2}+\frac {8 a^2}{x}+\frac {4 a^3}{i+a x}-12 i a^3 \log (x)+12 i a^3 \log (i+a x) \]

[Out]

-1/3/x^3-2*I*a/x^2+8*a^2/x+4*a^3/(I+a*x)-12*I*a^3*ln(x)+12*I*a^3*ln(I+a*x)

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5170, 90} \begin {gather*} \frac {4 a^3}{a x+i}-12 i a^3 \log (x)+12 i a^3 \log (a x+i)+\frac {8 a^2}{x}-\frac {2 i a}{x^2}-\frac {1}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^((4*I)*ArcTan[a*x])/x^4,x]

[Out]

-1/3*1/x^3 - ((2*I)*a)/x^2 + (8*a^2)/x + (4*a^3)/(I + a*x) - (12*I)*a^3*Log[x] + (12*I)*a^3*Log[I + a*x]

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 5170

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 - I*a*x)^(I*(n/2))/(1 + I*a*x)^(I*(n/2))
), x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[(I*n - 1)/2]

Rubi steps

\begin {align*} \int \frac {e^{4 i \tan ^{-1}(a x)}}{x^4} \, dx &=\int \frac {(1+i a x)^2}{x^4 (1-i a x)^2} \, dx\\ &=\int \left (\frac {1}{x^4}+\frac {4 i a}{x^3}-\frac {8 a^2}{x^2}-\frac {12 i a^3}{x}-\frac {4 a^4}{(i+a x)^2}+\frac {12 i a^4}{i+a x}\right ) \, dx\\ &=-\frac {1}{3 x^3}-\frac {2 i a}{x^2}+\frac {8 a^2}{x}+\frac {4 a^3}{i+a x}-12 i a^3 \log (x)+12 i a^3 \log (i+a x)\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.03, size = 62, normalized size = 1.00 \begin {gather*} -\frac {1}{3 x^3}-\frac {2 i a}{x^2}+\frac {8 a^2}{x}+\frac {4 a^3}{i+a x}-12 i a^3 \log (x)+12 i a^3 \log (i+a x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^((4*I)*ArcTan[a*x])/x^4,x]

[Out]

-1/3*1/x^3 - ((2*I)*a)/x^2 + (8*a^2)/x + (4*a^3)/(I + a*x) - (12*I)*a^3*Log[x] + (12*I)*a^3*Log[I + a*x]

________________________________________________________________________________________

Maple [A]
time = 0.10, size = 61, normalized size = 0.98

method	result	size
default	\(4 a^{4} \left (\frac {1}{a \left (a x +i\right )}+\frac {3 i \ln \left (a x +i\right )}{a}\right )-\frac {1}{3 x^{3}}-12 i a^{3} \ln \left (x \right )-\frac {2 i a}{x^{2}}+\frac {8 a^{2}}{x}\)	\(61\)
risch	\(\frac {12 a^{3} x^{3}+6 i a^{2} x^{2}+\frac {5}{3} a x -\frac {1}{3} i}{\left (a x +i\right ) x^{3}}+12 a^{3} \arctan \left (a x \right )+6 i a^{3} \ln \left (a^{2} x^{2}+1\right )-12 i a^{3} \ln \left (-x \right )\)	\(73\)
meijerg	\(\frac {a^{4} \left (-\frac {2 \left (-15 a^{4} x^{4}-10 a^{2} x^{2}+2\right )}{3 x^{3} \left (a^{2}\right )^{\frac {3}{2}} \left (2 a^{2} x^{2}+2\right )}+\frac {5 a^{3} \arctan \left (a x \right )}{\left (a^{2}\right )^{\frac {3}{2}}}\right )}{2 \sqrt {a^{2}}}+2 i a^{3} \left (\frac {3 a^{2} x^{2}}{3 a^{2} x^{2}+3}+2 \ln \left (a^{2} x^{2}+1\right )-1-4 \ln \left (x \right )-2 \ln \left (a^{2}\right )-\frac {1}{a^{2} x^{2}}\right )-\frac {3 a^{4} \left (-\frac {2 \left (3 a^{2} x^{2}+2\right )}{x \sqrt {a^{2}}\, \left (2 a^{2} x^{2}+2\right )}-\frac {3 a \arctan \left (a x \right )}{\sqrt {a^{2}}}\right )}{\sqrt {a^{2}}}-2 i a^{3} \left (-\frac {2 a^{2} x^{2}}{2 a^{2} x^{2}+2}-\ln \left (a^{2} x^{2}+1\right )+1+2 \ln \left (x \right )+\ln \left (a^{2}\right )\right )+\frac {a^{4} \left (\frac {2 x \sqrt {a^{2}}}{2 a^{2} x^{2}+2}+\frac {\sqrt {a^{2}}\, \arctan \left (a x \right )}{a}\right )}{2 \sqrt {a^{2}}}\)	\(272\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+I*a*x)^4/(a^2*x^2+1)^2/x^4,x,method=_RETURNVERBOSE)

[Out]

4*a^4*(1/a/(I+a*x)+3*I*ln(I+a*x)/a)-1/3/x^3-12*I*a^3*ln(x)-2*I*a/x^2+8*a^2/x

________________________________________________________________________________________

Maxima [A]
time = 0.48, size = 77, normalized size = 1.24 \begin {gather*} 12 \, a^{3} \arctan \left (a x\right ) + 6 i \, a^{3} \log \left (a^{2} x^{2} + 1\right ) - 12 i \, a^{3} \log \left (x\right ) + \frac {36 \, a^{4} x^{4} - 18 i \, a^{3} x^{3} + 23 \, a^{2} x^{2} - 6 i \, a x - 1}{3 \, {\left (a^{2} x^{5} + x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2/x^4,x, algorithm="maxima")

[Out]

12*a^3*arctan(a*x) + 6*I*a^3*log(a^2*x^2 + 1) - 12*I*a^3*log(x) + 1/3*(36*a^4*x^4 - 18*I*a^3*x^3 + 23*a^2*x^2
- 6*I*a*x - 1)/(a^2*x^5 + x^3)

________________________________________________________________________________________

Fricas [A]
time = 3.12, size = 86, normalized size = 1.39 \begin {gather*} \frac {36 \, a^{3} x^{3} + 18 i \, a^{2} x^{2} + 5 \, a x - 36 \, {\left (i \, a^{4} x^{4} - a^{3} x^{3}\right )} \log \left (x\right ) - 36 \, {\left (-i \, a^{4} x^{4} + a^{3} x^{3}\right )} \log \left (\frac {a x + i}{a}\right ) - i}{3 \, {\left (a x^{4} + i \, x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2/x^4,x, algorithm="fricas")

[Out]

1/3*(36*a^3*x^3 + 18*I*a^2*x^2 + 5*a*x - 36*(I*a^4*x^4 - a^3*x^3)*log(x) - 36*(-I*a^4*x^4 + a^3*x^3)*log((a*x
+ I)/a) - I)/(a*x^4 + I*x^3)

________________________________________________________________________________________

Sympy [A]
time = 0.20, size = 70, normalized size = 1.13 \begin {gather*} 12 a^{3} \left (- i \log {\left (24 a^{4} x \right )} + i \log {\left (24 a^{4} x + 24 i a^{3} \right )}\right ) + \frac {36 a^{3} x^{3} + 18 i a^{2} x^{2} + 5 a x - i}{3 a x^{4} + 3 i x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)**4/(a**2*x**2+1)**2/x**4,x)

[Out]

12*a**3*(-I*log(24*a**4*x) + I*log(24*a**4*x + 24*I*a**3)) + (36*a**3*x**3 + 18*I*a**2*x**2 + 5*a*x - I)/(3*a*
x**4 + 3*I*x**3)

________________________________________________________________________________________

Giac [A]
time = 0.41, size = 54, normalized size = 0.87 \begin {gather*} 12 i \, a^{3} \log \left (a x + i\right ) - 12 i \, a^{3} \log \left ({\left | x \right |}\right ) + \frac {36 \, a^{3} x^{3} + 18 i \, a^{2} x^{2} + 5 \, a x - i}{3 \, {\left (a x + i\right )} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2/x^4,x, algorithm="giac")

[Out]

12*I*a^3*log(a*x + I) - 12*I*a^3*log(abs(x)) + 1/3*(36*a^3*x^3 + 18*I*a^2*x^2 + 5*a*x - I)/((a*x + I)*x^3)

________________________________________________________________________________________

Mupad [B]
time = 0.13, size = 55, normalized size = 0.89 \begin {gather*} 24\,a^3\,\mathrm {atan}\left (2\,a\,x+1{}\mathrm {i}\right )+\frac {\frac {5\,x}{3}+12\,a^2\,x^3+a\,x^2\,6{}\mathrm {i}-\frac {1{}\mathrm {i}}{3\,a}}{x^4+\frac {x^3\,1{}\mathrm {i}}{a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x*1i + 1)^4/(x^4*(a^2*x^2 + 1)^2),x)

[Out]

24*a^3*atan(2*a*x + 1i) + ((5*x)/3 + a*x^2*6i - 1i/(3*a) + 12*a^2*x^3)/(x^4 + (x^3*1i)/a)

________________________________________________________________________________________